The present invention relates generally to an hydraulic drive for elevators and, in particular, to a method and apparatus for reducing the electrical energy required to drive an hydraulic drive pump in an elevator system.
To ascertain the power output requirements for the drive of an hydraulic elevator, a maximum carrying capacity and a speed of travel in the upward direction are selected for the elevator car. For a total mass "P" consisting of car weight, carrying capacity and possible additional weights and a predetermined desired travel speed "v" in the upward direction, a maximum power requirement "N" is calculated by the formula: EQU N=P.multidot.v
It is evident from this formula that the power requirement of an elevator drive is directly dependent on the mass to be moved and on the speed or elevator car.
However, the maximum power of the drive is utilized only when the elevator car is loaded to the full carrying capacity. For smaller loads, the drive operates with poorer efficiency and thus with relatively high energy consumption.
An hydraulic elevator having relatively low energy consumption is shown in the German patent document DE-OS 40 34 666. This elevator is driven by an inverter energy source and includes a main cylinder with a piston connected to an elevator car and an auxiliary or balancing cylinder with a fixed balancing weight and a selectable balancing weight. The cylinders are connected together by an hydraulic pump. The fixed balancing weight corresponds normally to the weight of the elevator car, while the selectable balancing weight corresponds to approximately one half the maximum load in the elevator car. Thereby, the pressure difference between the main cylinder and the balancing cylinder is kept as small as possible for all loads. The design of the size of the drive thus depends on the difference between the loading of the balancing cylinder and the main cylinder. The output power during downward travel without a load is calculated by the formula: EQU N=[(A+B+0)-(C+B+w/2)].multidot.v+(A-C-w/2).multidot.v
(when C=A): PA1 N=-(w/2).multidot.v PA1 (when C=A): PA1 N=+(w/2).multidot.v
The output power during downward travel at full load is calculated by the formula: EQU N=[(A+B+W)-(C+B+w/2)].multidot.v=(A-C+w/2).multidot.v
In each formula above, "N" is the output or driving power, "A" is the mass of the car, "B" is the mass of the piston, "W" is the mass of the maximum car load, "w/2" is the mass of one half of the car load, "C" is the mass of the fixed balancing weight (here C=A) and "v" is the lifting speed.
A disadvantage of this hydraulic elevator is that an additional auxiliary cylinder with a fixed and a selectable balancing weight, the necessary connecting line and a relatively expensive inverter energy source are required.